Question

16. Jar A has 4 red and 5 black candies. Jar B has 6 red and 2 black candies. A fair die is rolled, and jar A is selected if a number divisible by 3 comes up, otherwise, Jar B is selected. One candy is drawn from the jar.

a) What is the probability you selected Jar A and got a red candy?

b) What is the probability you selected Jar B and got a red candy?

c) What is the probability you got a red candy?

d) Suppose a red candy is drawn, what is the probability it came from jar A?

e) What is the probability Jar B was selected if a black candy is drawn?

Answer 1

16.

Number of red candies in Jar A = 4

Number of black candies in Jar A = 5

Total number of candies in Jar A =4+5 =9

Number of red candies in Jar B = 6

Number of black candies in Jar B = 2

Total number of candies in Jar B =6+2 =8

A : Event of selecting Jar A

B : Event of selecting Jar B

Jar A is selected if a number divisible by 3 come up when a die is rolled:

Numbers (in 1 to 6) divisible by 3 are : 3 and 6;

P(A) = Probability of selecting Jar A = Probability of a number divisible by 3 comes up = 2/6 =1/3

P(B) = 1-P(A) = 1-2/6 = 4/6 =2/3

a)

R : Event of getting a red candy

Probability you selected Jar A and got a red candy = P(A and R)

By multiplication theorem of probability

P(A and R)= P(A)P(R|A)

P(R|A) = Probability of getting a red candy given that Jar A is selected = Number of red candies in Jar A/Total number of candies in Jar A = 4/9

P(A) = 1/3

P(A and R)= P(A)P(R|A) = (1/3) x (4/9) = 4/27

Probability you selected Jar A and got a red candy = 4/27

b)

Probability you selected Jar B and got a red candy = P(B and R)

By multiplication theorem of probability

P(B and R)= P(B)P(R|B)

P(R|B) = Probability of getting a red candy given that Jar B is selected = Number of red candies in Jar B/Total number of candies in Jar B = 6/8 = 3/4

P(B) = 2/3

P(B and R)= P(B)P(R|B) = (2/3) x (3/4) = 1/2

Probability you selected Jar B and got a red candy = 1/2

c)

Probability you got a red candy = Probability (that you selected Jar A and got red candy OR that you selected Jar Band got red candy) = Probability you selected Jar A and got a red candy + Probability you selected Jar B and got a red candy = (4/27)+(1/2) = (8+27)54 = 35/54

Probability you got a red candy = 35/54

d)

Suppose red candy is drawn, Probability that it came from Jar A = P(A|R)

By Bayes theorem,

From (a): P(A)P(R|A) =  4/27

From (b): P(B)P(R|B) =  1/2

P(A)P(R|A)+P(B)P(R|B) = (4/27)+(1/2) = (8+27)54 = 35/54 (which was obtained in (c))

Suppose red candy is drawn, Probability that it came from Jar A = P(A|R) = 8/35

e)

Event of black candy is drawn= X

Probability Jar B was selected if Black candy is drawn = P(B|X)

By Bayes theorem

P(B) = 2/3

P(X|B) = Probability of getting a black candy given that Jar B is selected = Number of black candies in Jar B/ total number candies in Jar B = 2/8

P(B)P(X|B) = (2/3)(2/8) = 4/24 =1/6

P(A) = 1/3

P(X|A) = Probability of getting a black candy given that Jar A is selected = Number of black candies in Jar A/ total number candies in Jar A = 5/9

P(A)P(X|A) = (1/3)(5/9) = 5/27

P(A)P(X|A)+P(B)P(X|B) =(5/27) + (1/6) = 19/54

Probability Jar B was selected if Black candy is drawn = P(B|X) = 9/19

Probability Jar B was selected if Black candy is drawn = 9/19